Write $\angle ACB = \gamma$. Then $\angle ACF = \angle FCB = \frac{\gamma}{2}$, since $CF$ is the bisector of the angle $\angle ACB$. The points $A$, $D$, $E$ and $C$ are concyclic, so $\angle DAE = \angle DCE = \frac{\gamma}{2}$. The line $BF$ is parallel to the line $AE$, so $\angle ABF = \angle DAE = \frac{\gamma}{2}$. We have $\angle ACF = \frac{\gamma}{2} = \angle ABF$, so the points $A$, $F$, $B$ and $C$ are concyclic. From here we get $\angle FAB = \angle FCB = \frac{\gamma}{2} = \angle ABF$ and so $AFB$ is an isosceles triangle with the apex at $F$.